CMI-2019-DataScience-A: 6

Question

Let $R$ denote the set of real numbers and let $A=\{x\in R:x\neq 3\}$. For $x\in A$, let $f(x)=\frac{2x+1}{x-3}.$ Let $B$ denote the range of $f$. Then

• A. $B=\{x\in R:x \neq -2\} \;and \;f^{-1}(x)=\frac{3x-1}{x+2};$
• B. $B=\{x\in R:x \neq 2\}\; and\; f^{-1}(x)=\frac{3x+1}{x-2};$
• C. $B=\{x\in R:x \neq 2\}\; and\; f^{-1}(x)=\frac{3x-1}{x-2};$
• D. $f^{-1}(x)$ does not exist because $f$ is not injective.

Answer 1

Range of f(x) can be expressed as the domain of the inverse of f(x)...!!

Now, inverse of the function f(x) can be found as...

y = (2x + 1) / (x - 3)
x.y - 3.y = 2.x + 1
x.(y - 2) = 1 + 3.y
x = (1 + 3y) / (y - 2)...!!
So, x = f(y) becomes our function as the inverse of f(x)...!!
And so, the domain of this function (required range) is, all the real numbers (R) (except (2))...!!
So, OPTION B is correct..!!